Problem: A bag contains $8$ red balls, $7$ green balls, and $6$ blue balls. If a ball is randomly chosen, what is the probability that it is not green?
Solution: There are $8 + 7 + 6 = 21$ balls in the bag. There are $7$ green balls. That means $21 - 7 = 14$ are not green. The probability is $ \frac{14}{21} = \dfrac{2}{3}$.